Differentiating under the integral sign
Start from the integral formula and differentiate with respect to
a. The contour and f(z) do not depend on
a, so only 1/(z - a) is touched.
Step 1 — differentiate the kernel once. Since
\dfrac{\partial}{\partial a}\dfrac{1}{z - a} = \dfrac{1}{(z - a)^2}:
f'(a) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^2}\,dz.
Step 2 — differentiate again. Each derivative drops the power by one and
multiplies by the old power:
f''(a) = \frac{2!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^3}\,dz.
Step 3 — the general pattern. Differentiating
n times brings out a factor of n! and
raises the denominator to n + 1:
f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^{\,n+1}}\,dz.
The integral on the right exists for every n, so every
derivative exists. A function that is complex-differentiable once is automatically smooth to all
orders.
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If f is holomorphic, then it has derivatives of all orders, given
by \displaystyle f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^{\,n+1}}\,dz.
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One complex derivative forces infinitely many — utterly unlike real calculus, where a
once-differentiable function may have no second derivative at all.
This is one of complex analysis's most shocking departures from real calculus. On the real
line, functions that are differentiable once but not twice are commonplace — the tidy
example is x^2\sin(1/x) (patched to 0 at
the origin), which has a first derivative everywhere but no second derivative at
0. You can freely build a real function differentiable exactly
k times and no more.
In the complex world that whole ladder collapses. The complex derivative is so rigid — through
Cauchy's integral formula the value of every
f^{(n)}(a) is already pinned down by a contour integral of
f alone — that a single derivative forces the entire infinite
cascade. So never go looking for a holomorphic function that is "differentiable exactly twice":
it cannot exist. One implies all, or none at all.
Liouville's theorem: bounded entire ⟹ constant
A function holomorphic on the whole plane is called entire. The
n = 1 formula gives a clean bound on its derivative, and that bound
proves a striking rigidity.
Step 1 — bound f' on a big circle. Take
\gamma to be the circle of radius R about
a, and suppose |f| \le M everywhere. On
that circle |z - a| = R, so the integrand of
f' has size at most M/R^2. The
ML
inequality with L = 2\pi R gives
|f'(a)| \le \frac{1}{2\pi}\cdot\frac{M}{R^2}\cdot 2\pi R = \frac{M}{R}.
Step 2 — let the radius grow. Because f is entire,
this holds for every R. Send
R \to \infty:
|f'(a)| \le \frac{M}{R} \;\longrightarrow\; 0.
Step 3 — conclude. So f'(a) = 0 at every point
a. A function with zero derivative everywhere is constant.
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A bounded entire function is constant.
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The proof is one ML estimate, |f'(a)| \le M/R, pushed to the limit
R \to \infty.
Liouville's theorem cracks a classic problem in one stroke. Suppose a non-constant polynomial
p(z) had no root. Then
1/p(z) would be holomorphic on the whole plane — entire. As
|z| \to \infty a polynomial grows without bound, so
1/p(z) \to 0; combined with continuity, that makes
1/p bounded. A bounded entire function is constant by
Liouville, so 1/p — and hence p — would
be constant, contradicting the assumption. Therefore every non-constant polynomial has a root:
the fundamental
theorem of algebra, delivered by complex integration.